Question

Find $\frac{dy}{dx}$ for $y=$${\sin }^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right)$

Answer 1

$y=si{n}^{-1}\left[\frac{3x+4\sqrt{1-x^2}}{5}\right]$

put $x=cos\theta$

$y=si{n}^{-1}\left[\frac{3cos\theta +4\sqrt{1-co{s}^2\theta }}{5}\right]$

$y=si{n}^{-1}[\frac{3}{5}cos\theta +\frac{4}{5}\sqrt{si{n}^2\theta }]$

$y=si{n}^{-1}[\frac{4}{5}in\theta +\frac{3}{5}cos\theta ]$

$sin\alpha =\frac{4}{5};cos\alpha =\frac{3}{5}$

$y=si{n}^{-1}[sin\alpha sin\theta +cos\alpha cos\theta ]$

$=si{n}^{-1}\left[sin\right(\theta -\alpha \left)\right]$

$y=\theta -\alpha$

$y=co{s}^{-1}x-\alpha$ $[\because \alpha$ is constant]

$\frac{d}{dx}y=\frac{-1}{\sqrt{1-x^2}}$